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 A problem on factors (Posted on 2007-07-16)
Let P(X,A) be the number of positive divisors of A not greater than X; for example, P(1,12)=1, P(2,12)=2, and P(12,12)=6.

1)If P(12,A)=7 and P(A,A)=10, what's A?
2)If x*y is a divisor of A, then show that P(x*y,A) ≥ P(x,A)+P(y,A)-1

 See The Solution Submitted by Praneeth Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Part 1 | Comment 1 of 2
Since we know A has 10 positive divisors, we know it is of the form p^4*q^1 for primes p and q.  Since 7 of these divisors are 12 or less it makes sense that p and q be relatively small.  Trying 2^4*3^1=48 works perfectly.

P(12,48)=7 {1,2,4,8,3,6,12}
P(48,48)=10 {1,2,4,8,16,3,6,12,24,48}

This P is an interesting function.  I think a harder version of part 1 would be nice to see.

I haven't started part 2 although its obviously true if x and y are primes and x*y=A.  4=3+2-1

Edited on July 17, 2007, 9:40 am

Edited on July 17, 2007, 9:41 am
 Posted by Jer on 2007-07-16 10:57:36

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